
In a transistor, doping level in base is increased slightly. How will it affect
(a) Collector current and (b) base current
Answer
494.4k+ views
Hint: In order to solve this question we need to understand transistors. So a transistor is a combination of doped semiconductors. There are two types of transistors one is PNP and other is NPN and three regions known as Emitter, Base and Collector respectively. These three regions are named according to their role and we do several biases in order to use a transistor like amplifier etc. Also transistors can work in three particular regions known as cutoff, active and saturation state.
Complete answer:
Before we understand what happens if doping is increased let us understand how transistors work. Consider a NPN transistor forward bias. So in base (P) we have holes as a majority charge carrier so due to applied electric field from base to emitter holes in base and electrons in emitter move towards opposite end and constitute Emitter current (Emanating outside from Emitter).
Now due to electron and hole movement some electrons and holes recombine with each other in base region and due to the same electrons number kept decreasing in base so battery provides holes in p side or to base so base current constitutes entering in base. While at collector base junction minority charge carrier movement increases due to reverse biasing of junction and hence collector current constitutes entering inside collector.
${I_E} = {I_B} + {I_C}$
This relation is always true
Now if doping increases then at the emitter base junction majority charges increase so their movement increases and hence there is increase in Emitter current.
(a) While at collector-base junction minority charges increase so their movement increases and hence collector current increases.
(b) Due to increase in doping both holes and electrons increase so their recombination in base region would not affect so battery needs to provide less holes and hence base current decreases.
Note: It should be remembered that increases doping however increases emitter and collector current so it is indirectly increasing reverse saturation current at collector-base terminal so avalanche breakdown could happen and diode may be useless so to prevent this something extra is done like we mix the impurity at collector side to some extent so to decrease this case.
Complete answer:
Before we understand what happens if doping is increased let us understand how transistors work. Consider a NPN transistor forward bias. So in base (P) we have holes as a majority charge carrier so due to applied electric field from base to emitter holes in base and electrons in emitter move towards opposite end and constitute Emitter current (Emanating outside from Emitter).
Now due to electron and hole movement some electrons and holes recombine with each other in base region and due to the same electrons number kept decreasing in base so battery provides holes in p side or to base so base current constitutes entering in base. While at collector base junction minority charge carrier movement increases due to reverse biasing of junction and hence collector current constitutes entering inside collector.
${I_E} = {I_B} + {I_C}$
This relation is always true
Now if doping increases then at the emitter base junction majority charges increase so their movement increases and hence there is increase in Emitter current.
(a) While at collector-base junction minority charges increase so their movement increases and hence collector current increases.
(b) Due to increase in doping both holes and electrons increase so their recombination in base region would not affect so battery needs to provide less holes and hence base current decreases.
Note: It should be remembered that increases doping however increases emitter and collector current so it is indirectly increasing reverse saturation current at collector-base terminal so avalanche breakdown could happen and diode may be useless so to prevent this something extra is done like we mix the impurity at collector side to some extent so to decrease this case.
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